\(\int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx\) [169]

Optimal result

Integrand size = 28, antiderivative size = 227 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx=\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}-\frac {\left (\frac {5}{4}+\frac {3 i}{4}\right ) \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a d^{3/2} f}+\frac {\left (\frac {5}{4}-\frac {3 i}{4}\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a d^{3/2} f}-\frac {5}{2 a d f \sqrt {d \tan (e+f x)}}+\frac {1}{2 d f \sqrt {d \tan (e+f x)} (a+i a \tan (e+f x))} \] Output:

(5/8+3/8*I)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a/d^(3/ 
2)/f-(5/8+3/8*I)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/2)/a/ 
d^(3/2)/f+(5/8-3/8*I)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1/2)/(d^(1/2)+d^(1/2 
)*tan(f*x+e)))*2^(1/2)/a/d^(3/2)/f-5/2/a/d/f/(d*tan(f*x+e))^(1/2)+1/2/d/f/ 
(d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.44 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx=\frac {-i-4 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-i \tan (e+f x)\right ) (-i+\tan (e+f x))-\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},i \tan (e+f x)\right ) (-i+\tan (e+f x))}{2 a d f \sqrt {d \tan (e+f x)} (-i+\tan (e+f x))} \] Input:

Integrate[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])),x]
 

Output:

(-I - 4*Hypergeometric2F1[-1/2, 1, 1/2, (-I)*Tan[e + f*x]]*(-I + Tan[e + f 
*x]) - Hypergeometric2F1[-1/2, 1, 1/2, I*Tan[e + f*x]]*(-I + Tan[e + f*x]) 
)/(2*a*d*f*Sqrt[d*Tan[e + f*x]]*(-I + Tan[e + f*x]))
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.17, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.607, Rules used = {3042, 4035, 27, 3042, 4012, 25, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (d \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4035

\(\displaystyle \frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}-\frac {\int -\frac {5 a d-3 i a d \tan (e+f x)}{2 (d \tan (e+f x))^{3/2}}dx}{2 a^2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5 a d-3 i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 a d-3 i a d \tan (e+f x)}{(d \tan (e+f x))^{3/2}}dx}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}+\frac {\int -\frac {3 i a d^2+5 a \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {3 i a d^2+5 a \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {\int \frac {3 i a d^2+5 a \tan (e+f x) d^2}{\sqrt {d \tan (e+f x)}}dx}{d^2}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 \int \frac {a d^2 (5 \tan (e+f x) d+3 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{d^2 f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \int \frac {5 \tan (e+f x) d+3 i d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {-\frac {10 a}{f \sqrt {d \tan (e+f x)}}-\frac {2 a \left (\left (\frac {5}{2}+\frac {3 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {5}{2}-\frac {3 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{4 a^2 d}+\frac {1}{2 d f (a+i a \tan (e+f x)) \sqrt {d \tan (e+f x)}}\)

Input:

Int[1/((d*Tan[e + f*x])^(3/2)*(a + I*a*Tan[e + f*x])),x]
 

Output:

1/(2*d*f*Sqrt[d*Tan[e + f*x]]*(a + I*a*Tan[e + f*x])) + ((-2*a*((5/2 + (3* 
I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[ 
d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d] 
)) - (5/2 - (3*I)/2)*(-1/2*Log[d + d*Tan[e + f*x] - Sqrt[2]*Sqrt[d]*Sqrt[d 
*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[2]*Sqrt[ 
d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]))))/f - (10*a)/(f*Sqrt[d*Tan[e 
 + f*x]]))/(4*a^2*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4012

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ 
(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2)   Int[(a + b*Tan[e + f*x] 
)^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 
]
 

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4035

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-a)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b* 
c - a*d)*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d))   Int[(c + d 
*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
&& NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0]
 

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.51

method	result	size
derivativedivides	\(\frac {2 d^{2} \left (-\frac {1}{d^{3} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{d \tan \left (f x +e \right )-i d}-\frac {4 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{3} \sqrt {i d}}\right )}{f a}\)	\(116\)
default	\(\frac {2 d^{2} \left (-\frac {1}{d^{3} \sqrt {d \tan \left (f x +e \right )}}+\frac {-\frac {\sqrt {d \tan \left (f x +e \right )}}{d \tan \left (f x +e \right )-i d}-\frac {4 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{\sqrt {-i d}}}{4 d^{3}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{4 d^{3} \sqrt {i d}}\right )}{f a}\)	\(116\)

Input:

int(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

2/f/a*d^2*(-1/d^3/(d*tan(f*x+e))^(1/2)+1/4/d^3*(-(d*tan(f*x+e))^(1/2)/(d*t 
an(f*x+e)-I*d)-4/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^(1/2)))-1 
/4/d^3/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(1/2)))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 645 vs. \(2 (167) = 334\).

Time = 0.12 (sec) , antiderivative size = 645, normalized size of antiderivative = 2.84 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx =\text {Too large to display} \] Input:

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e)),x, algorithm="fricas")
 

Output:

1/4*((a*d^2*f*e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e))*sqrt(1/4* 
I/(a^2*d^3*f^2))*log(-2*(2*(I*a*d^2*f*e^(2*I*f*x + 2*I*e) + I*a*d^2*f)*sqr 
t((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/4*I/( 
a^2*d^3*f^2)) + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) - (a*d^2*f* 
e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e))*sqrt(1/4*I/(a^2*d^3*f^2 
))*log(-2*(2*(-I*a*d^2*f*e^(2*I*f*x + 2*I*e) - I*a*d^2*f)*sqrt((-I*d*e^(2* 
I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/4*I/(a^2*d^3*f^2)) 
 + I*d*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)) + (a*d^2*f*e^(4*I*f*x + 
4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e))*sqrt(-4*I/(a^2*d^3*f^2))*log(((a*d*f 
*e^(2*I*f*x + 2*I*e) + a*d*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2* 
I*f*x + 2*I*e) + 1))*sqrt(-4*I/(a^2*d^3*f^2)) + 2)*e^(-2*I*f*x - 2*I*e)/(a 
*d*f)) - (a*d^2*f*e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e))*sqrt( 
-4*I/(a^2*d^3*f^2))*log(-((a*d*f*e^(2*I*f*x + 2*I*e) + a*d*f)*sqrt((-I*d*e 
^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-4*I/(a^2*d^3*f^ 
2)) - 2)*e^(-2*I*f*x - 2*I*e)/(a*d*f)) + sqrt((-I*d*e^(2*I*f*x + 2*I*e) + 
I*d)/(e^(2*I*f*x + 2*I*e) + 1))*(-9*I*e^(4*I*f*x + 4*I*e) - 8*I*e^(2*I*f*x 
 + 2*I*e) + I))/(a*d^2*f*e^(4*I*f*x + 4*I*e) - a*d^2*f*e^(2*I*f*x + 2*I*e) 
)
 

Sympy [F]

\[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx=- \frac {i \int \frac {1}{\left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )} - i \left (d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx}{a} \] Input:

integrate(1/(d*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e)),x)
 

Output:

-I*Integral(1/((d*tan(e + f*x))**(3/2)*tan(e + f*x) - I*(d*tan(e + f*x))** 
(3/2)), x)/a
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e)),x, algorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx=-\frac {i \, {\left (\frac {-5 i \, d \tan \left (f x + e\right ) - 4 \, d}{\sqrt {d \tan \left (f x + e\right )} d \tan \left (f x + e\right ) - i \, \sqrt {d \tan \left (f x + e\right )} d} - \frac {i \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\sqrt {d} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} - \frac {4 i \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\sqrt {d} {\left (-\frac {i \, d}{{\left | d \right |}} + 1\right )}}\right )}}{2 \, a d f} \] Input:

integrate(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e)),x, algorithm="giac")
 

Output:

-1/2*I*((-5*I*d*tan(f*x + e) - 4*d)/(sqrt(d*tan(f*x + e))*d*tan(f*x + e) - 
 I*sqrt(d*tan(f*x + e))*d) - I*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d 
)/(I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(sqrt(d)*(I*d/abs(d) + 1)) 
 - 4*I*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + 
sqrt(2)*sqrt(d)*abs(d)))/(sqrt(d)*(-I*d/abs(d) + 1)))/(a*d*f)
 

Mupad [B] (verification not implemented)

Time = 3.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.65 \[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx=-\frac {-\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{2\,a\,f}+\frac {2{}\mathrm {i}}{a\,f}}{-{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}+d\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,1{}\mathrm {i}}+2\,\mathrm {atanh}\left (a\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {1{}\mathrm {i}}{a^2\,d^3\,f^2}}\right )\,\sqrt {-\frac {1{}\mathrm {i}}{a^2\,d^3\,f^2}}+2\,\mathrm {atanh}\left (4\,a\,d\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {1{}\mathrm {i}}{16\,a^2\,d^3\,f^2}}\right )\,\sqrt {\frac {1{}\mathrm {i}}{16\,a^2\,d^3\,f^2}} \] Input:

int(1/((d*tan(e + f*x))^(3/2)*(a + a*tan(e + f*x)*1i)),x)
 

Output:

2*atanh(a*d*f*(d*tan(e + f*x))^(1/2)*(-1i/(a^2*d^3*f^2))^(1/2))*(-1i/(a^2* 
d^3*f^2))^(1/2) - (2i/(a*f) - (5*tan(e + f*x))/(2*a*f))/(d*(d*tan(e + f*x) 
)^(1/2)*1i - (d*tan(e + f*x))^(3/2)) + 2*atanh(4*a*d*f*(d*tan(e + f*x))^(1 
/2)*(1i/(16*a^2*d^3*f^2))^(1/2))*(1i/(16*a^2*d^3*f^2))^(1/2)
 

Reduce [F]

\[ \int \frac {1}{(d \tan (e+f x))^{3/2} (a+i a \tan (e+f x))} \, dx=\frac {\sqrt {d}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}}{\tan \left (f x +e \right )^{3} i +\tan \left (f x +e \right )^{2}}d x \right )}{a \,d^{2}} \] Input:

int(1/(d*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e)),x)
 

Output:

(sqrt(d)*int(sqrt(tan(e + f*x))/(tan(e + f*x)**3*i + tan(e + f*x)**2),x))/ 
(a*d**2)